so we finished the last video with a suggestion about how we might evaluate a definite integral avoiding the whole limit of a Riemann sum process and so we talked about the first fundamental theorem and we worked with the idea that the first fundamental theorem allows us which is the idea of finding the derivative of these particular types of functions involving definite integrals where we have a variable in one of the limits of integration and some of the different things that we could do given that fact of that function being an antiderivative and knowing how to find its derivative and then we finished with the example where we were starting to see where this might lead us so this is where we'll introduce the second part of the fundamental theorem of calculus or the fundamental theorem of calculus part two I prefer to call it the fundamental theorem of integral calculus because it really drives everything that we do in integral calculus and so let's go ahead and state that so if we have a continuous function lowercase F on an interval A to B and that interval is closed so we want to be continuous throughout and at the endpoints and suppose that capital F is an antiderivative for lowercase F on the same interval so lowercase F is our integrand capital F is the antiderivative of our integrand and so the result we have is that the definite integral from A to B of f of X DX is equal to the antiderivative evaluated at B subtract the antiderivative evaluated at a so we evaluating this definite integral again remember what this ties back to you this ties back to the limit of a Riemann sum this definite integral has this definition right the limit of the Riemann sum so we don't have to worry about the fact that we need that this is a limit of a Riemann sum underneath the hood now we want to keep that in mind this ties back to the limit of a Riemann sum but it turns out that all we need to evaluate this limit and consequently this definite integral is to find the antiderivative and then plug in the upper limit subtract and then plug in the lower limit so we're not going to be stuck with having to work to that clunky cumbersome limit anymore all we need are two numbers we need these numbers capital F of B and capital F of a and we subtract and so that's all it takes to evaluate the definite integral is just to find the antiderivative evaluate at the upper limit subtract evaluate at the lower limit now the trade-off is apparent hopefully finding the antiderivative can often be difficult for us this semester the antiderivatives we will be working with towards finding are not going to be too difficult at the end of the day if you're moving forward into math 1080 you'll start to get a good dose of integration techniques where the problem is finding the antiderivative and there's a different collection of techniques and tools that we need in order to find the antiderivative sometimes for us this semester we're really not going to get too deep into the woods there so to evaluate a definite integral we've got a notation so the notation we'll use is well write our antiderivative we'll write this bar with the lower limit in the upper limit and then we'll do our evaluation so since we're subtracting remember remember the antiderivative has a plus C attached but we're subtracting so for this part we'll have + C and this part we'll have minus C so that C minus C is always going to cancel so when we're finding a definite integral we can know that we can ignore the plus C doesn't need to be written so let's go back to that question we started with how do we find the area under the graph of x squared plus 1 on the interval 0 to 2 so we worked through the limit of a Riemann sum using right-hand Point endpoints and a uniform partition and we found that limit was equal to 14 thirds consequently the area under the curve was 14 third square units so now we can just evaluate this directly we're going to evaluate the definite integral from 0 to 2 of x squared plus 1 with respect to X so the fundamental theorem says find the antiderivative our antiderivative is going to be 1/3 X cubed plus X and then we're going to evaluate at the endpoints or the limits of integration 2 and 0 so the first thing the upper limit goes in first so 2 is going to be plugged into the antiderivative so that's going to be 1/3 of 2 cubed which is 8 thirds plus 2 and then we'll subtract plug in the lower limit while plugging in 0 for both is just going to be 0 all right so 8/3 plus 2 2 is going to be 6 thirds and so there is the 14 thirds that we worked so hard for previously and so this is how we are going to approach the evaluation of definite integrals now so all we need to do is find the antiderivative plug in the upper limit subtract plug in the lower limit quite remarkable that that limit turns out to be just a simple evaluation of the antiderivative at two points and then subtracting so let's go through a handful of these and get some practice let's look at the definite integral from negative 1 to 3 of 2 plus X so we're going to find the antiderivative the antiderivative of 2 is 2x the antiderivative of X is going to be 1/2 x squared so remember we're reversing the power rule so we're increasing the power by 1 dividing by the new power and we're going to evaluate at our endpoints or our bounds negative 1 and 3 so 3 goes in first that's going to be 2 times 3 plus 1/2 times 3 squared that's going to be 9 ABS we're going to subtract now negative 1 goes in that's going to be negative 2 plus 1/2 negative one squared is of course positive one so let's go ahead and just write six plus nine halves so that negative distributes we'll get plus two minus one-half all right six plus two is of course eight nine halves minus one halves is eight halves eight halves is of course four so we have eight and four together giving us a final result here of twelve all right next up the definite integral from negative 2 to negative 1/2 over 3x DX this is the same as 2/3 we'll bring the constant out negative 2 to negative 1 and then I just have 1 over X DX alright you'll remember the antiderivative for 1 over X is the natural log of the absolute value of x and then we will evaluate at our bounds negative 2 and negative 1 now here's why the absolute value is important for this function there's no restriction that we cannot plug in negative 2 negative 1 the only thing we can't plug into this function is 0 however if I were to leave the absolute value off we would be trying to calculate in the antiderivative the natural log of negative 1 and the natural log of negative 2 well the net natural log is undefined for 0 and less so we cannot finish this without the absolute value so we'll apply the fundamental theorem so we'll have 2/3 natural log absolute value of negative 1 that will be positive 1 and then minus 2/3 natural log absolute value of negative 2 which is of course positive to you well the natural log of 1 is 0 so this integral turns out to just be equal to negative 2/3 natural log 2 right here's the definite integral from PI over 3 2 PI over 4 of the secant of X excuse me the secant squared of X DX well the antiderivative of secant is the tangent and then we'll just simply evaluate at PI over 3 and PI over 4 so this becomes the tangent of the upper limits minus the tangent of the lower limit well we know the tangent of PI over 4 is 1 the tangent of PI over 3 you will find to be the square root of 3 so this turns out to be 1 minus root 3 now here we have the definite integral from 0 to 1,000 now what's here is understood to just be a 1 so very simply our antiderivative is just X and then we have the fundamental theorem evaluate at the upper limit subtract evaluate at the lower limit right so this is just equal to a thousand and so what you might see here if you were to graph this this is just the function alright certainly not to scale here but this is just a rectangle and we're finding the area of that rectangle right width times height 1,000 times 1 all right here's a definite integral from 4 to 9 I have the square root of U cubed minus 1 divided by u cubed D u so what we'd like to do here we have a quotient we should simplify this as best we can so I have u cubed over u cubed in the square root minus 1 over u cubed well u cubed over u cubed under a root is just equal to 1 and then now this is 1 over the square root of U cubed so that's going to be minus U to the negative three-halves all right so our antiderivative the antiderivative of 1 with respect to you it's just going to be you the antiderivative of U to the negative three-halves we're going to add one to the power so negative three-halves plus two over two that's going to be negative one half divided by the new power of negative one half all right and then we'll evaluate at the endpoints four and nine now let's simplify this just to save myself some room here we've got a double negative so that's going to become positive now this over 1/2 is going to be the same as times two and then you to the negative 1/2 is going to be root u on the bottom so here's our antiderivative just reexpress in an equivalent way so we'll evaluate using the fundamental theorem so nine goes in first so that'll be 9 plus 2 over root 9 which is of course 3 and then we'll subtract 4 plus 2 over root 4 which is going to be 2 all right this negative will distribute so we get minus 4 minus 2 over 2 which is of course just one so we're going to have 9 minus 4 minus 1 that's going to be 4 and then we're going to have 4 plus 2/3 and 4 plus 2/3 is going to be 14 thirds all right next up we have a definite integral from 0 to PI over 2 my integrand is 2 minus 3 sine theta d-theta so our antiderivative is going to be 2 theta and how do we get negative 3 sine theta well that would be the derivative of 3 cosine theta and then we'll evaluate at 0 and PI over 2 so you know one way of checking yourself take the derivative of this expression you should have the integrand so the derivative of 2 theta is 2 the derivative of 3 cosine theta will be negative three sine theta so we're in good shape so we'll plug in PI over two so we'll have two times pi over two plus three times the cosine of PI over two and then we'll subtract we'll plug in zero we'll have two times zero plus three times the cosine of zero all right well cosine PI over 2 is 0 so this term will drop cosine of 0 is 1 and then these twos will cancel so what do I have I'm going to have PI plus 0 minus 0 minus 1 so this integral evaluates to PI minus 1 all right so this all this process all centers on finding the antiderivative and then the fundamental theorem says that we plug in the upper limit subtract plug in the lower limit that's all there is to it here's the definite integral from 1 to root 3 of 4 over 1 plus x squared so hopefully you recognize the antiderivative is gonna be for our tangent of X and then we'll evaluate at our bounds of root 3 and 1 so this is for arctangent root 3 minus 4 arctangent 1 so arctangent of root 3 so what's the angle with tangent root 3 that's gonna be PI over 3 and what's the angle with tangent 1 that's gonna be PI over 4 so this is going to come out to 4 PI over 3 minus PI after these 4s cancel all right a couple less problems from this section we can use this knowledge now of the fundamental theorem of calculus to solve integral type equations so what I want to do here is I want to find the value that should go in the upper limit to make this integral evaluate to negative 4 so I want to solve the equation the definite integral from 0 to X so 14 minus 9 DT equal to negative or so what limit or limits of integration need to go here in order to make this evaluate to negative four so what we have we have the definite integral from 0 to X of 40 minus 9 DT so the fundamental theorem tells us how to evaluate this we'll find the antiderivative so the antiderivative of 4t will be tan 2t squared the antiderivative of 9 will be 90 and then we'll plug in the endpoints of X and 0 so this becomes 2x squared minus 9x and then plugging in 0 we'll just have a 0 so this is just going to be equal to 2x squared minus 9x so what we want to happen we want to x squared minus 9x to equal negative 4 so now we're just solving for x like a like anything we've done before right nothing new at this point so to solve for X here I'm going to add that 4 over so we'll have 2x squared minus 9 X plus 4 equal to 0 and so this will factor this is going to factor into 2x minus 1 and then X minus 4 all right so we can check that there's 2x squared there's minus 8 X minus 1 X can be minus 9 X negative 1 negative 4 positive 4 so we're going to have two solutions here 2x minus 1 equals 0 meaning X is 1/2 and X minus 4 equals 0 meaning X is 4 all right so there are two particular numbers here that we could put in this upper limit of integration in order to force this integral to evaluate to negative 4 we could put 1/2 and we could also put positive 4 and along the same lines let's do one more of these so if X is larger than zero let's solve for X so I want to find the values of x that will force this integral to equal pi so the integral X to x squared of 1 over T DT and I'd like that to evaluate to PI so again fundamental theorem tells me how to start this so the integral definite integral from X to x squared of 1 over T DT the antiderivative is the natural log of the absolute value of T fundamental theorem evaluate at the endpoints so we're gonna have the natural log of now we've set that x is positive so I can go ahead and neglect and we know we can an elected absolute value for the x squared but we can go ahead and neglect the absolute value we'll write the natural log of x squared minus the natural log of X right just applying the fundamental theorem right let's simplify this properties of logarithms that power of two can come out so I have two Ln x minus Ln X so if I - Ln X take away an Ln X just have an Ln X left over so what do we want to happen so we want this integral to equal pi in other words we want the natural log of X which is equal to this integral to equal pi so we want the natural log of X to equal pi right so exponentiating both sides we'll have our solution x is going to be equal to e to the PI all right so for this integral if I put e to the PI here and e to the PI squared here now I know that integral will evaluate to this result that I want